MCQ
$\int_{}^{} {{{\sec }^p}x\tan x\;dx = } $
- A$\frac{{{{\sec }^{p + 1}}x}}{{p + 1}} + c$
- ✓$\frac{{{{\sec }^p}x}}{p} + c$
- C$\frac{{{{\tan }^{p + 1}}x}}{{p + 1}} + c$
- D$\frac{{{{\tan }^p}x}}{p} + c$
therefore $\int_{}^{} {{{\sec }^p}x\tan x\,dx} = \int_{}^{} {{t^{p - 1}}dt = \frac{{{t^p}}}{p} + c} = \frac{{{{\sec }^p}x}}{p} + c.$
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